Question

man weighing 50kg is standing in lift find the thrust on floor of lift when lift ascends with acceleration of 5m/s^2. lift decends with an acceleration of 5m/s^2. lift ascending with uniform velocity of 3m/s​

Answer 1

Correct Question :-

▪ Man weighing 50 kg is standing in lift, find the thrust on floor of lift when lift ascends with acceleration of 5 m/s² and when it ascends with uniform initial velocity of 3 m/s² while going upward.

We have two situations here, so we would solve them one by one.

But first, We will take every physical quantity in the direction of motion positive and those not in direction of motion negative.

Case 1 : While going upward( ascending )

While ascending we are given the following values,

• a = + 5 m/s²
• g = - 10 m/s²

While going upward, there will be a pseudo force acting on the body in the downward direction where the acceleration is -5 m/s² [ acting in downward direction ]

⇒ Net force (Thrust) = mg + ma

⇒ Thrust = 50 × -10 - 50 × -5

⇒ Thrust = -500 - 250

⇒ Thrust = -750 newton.

Where, The magnitude of thrust is 750 newton and the negative sign here, signifies that it is acting in the opposite direction of motion. That's why we feel heavier while going upward in a lift.

Case 2 : While going upward ( u = 3 m/s² )

Here, the acceleration due to gravity is acting opposite to the direction of motion, so it would be taken negative.

Also it is given in the question that the velocity is uniform that means the acceleration, a is 0 m/s² .

So, We have

• g = - 10 m/s²
• a = 0 m/s²

Now,

⇒ Net Force (Thrust) = mg + ma

⇒ Thrust = 50×-10 + 50×0

⇒ Thrust = -500 newton

Here, The magnitude of thrust is 500 newton and the negative sign signifies that the thrust is in the opposite direction of motion.

Answer 2

Answer:

Correct Question :-

▪ Man weighing 50 kg is standing in lift, find the thrust on floor of lift when lift ascends with acceleration of 5 m/s² and when it ascends with uniform initial velocity of 3 m/s² while going upward.

We have two situations here, so we would solve them one by one.

But first, We will take every physical quantity in the direction of motion positive and those not in direction of motion negative.

Case 1 : While going upward( ascending )

While ascending we are given the following values,

a = + 5 m/s²

g = - 10 m/s²

While going upward, there will be a pseudo force acting on the body in the downward direction where the acceleration is -5 m/s² [ acting in downward direction ]

⇒ Net force (Thrust) = mg + ma

⇒ Thrust = 50 × -10 - 50 × -5

⇒ Thrust = -500 - 250

⇒ Thrust = -750 newton.

Where, The magnitude of thrust is 750 newton and the negative sign here, signifies that it is acting in the opposite direction of motion. That's why we feel heavier while going upward in a lift.

Case 2 : While going upward ( u = 3 m/s² )

Here, the acceleration due to gravity is acting opposite to the direction of motion, so it would be taken negative.

Also it is given in the question that the velocity is uniform that means the acceleration, a is 0 m/s² .

So, We have

g = - 10 m/s²

a = 0 m/s²

Now,

⇒ Net Force (Thrust) = mg + ma

⇒ Thrust = 50×-10 + 50×0

⇒ Thrust = -500 newton

Here, The magnitude of thrust is 500 newton and the negative sign signifies that the thrust is in the opposite direction of motion.

Explanation:

Hope this answer will help you.